Q83 / 160 · TS EAMCET 2015 · Mathematics
If $x=1+\frac{3}{1 !} \times \frac{1}{6}+\frac{3 \times 7}{2 !}\left(\frac{1}{6}\right)^2$ $+\frac{3 \times 7 \times 11}{3 !}\left(\frac{1}{6}\right)^3+\ldots$, then $x^4$ equals
- A81
- B54
- Correct answerC27
- D8
Solution
$$
\begin{aligned}
& \because x=1+\frac{3}{1 !} \times \frac{1}{6}+\frac{3 \times 7}{2 !}\left(\frac{1}{6}\right)^2 \\
&+\frac{3 \times 7 \times 11}{3 !}\left(\frac{1}{6}\right)^3+\ldots \ldots \\
& \therefore \quad(1-\alpha)^{-p / q} \\
&= 1+\left(\frac{p}{q}\right) \frac{1}{1 !}(\alpha)+\frac{\frac{p}{q}\left(\frac{p}{q}+1\right)}{2 !} \alpha^2 \\
&+\frac{\frac{p}{q}\left(\frac{p}{q}+1\right)\left(\frac{p}{q}+2\right)}{3 !} \alpha^3+\ldots \\
&=1+\frac{p}{1 !}\left(\frac{\alpha}{q}\right)+\frac{p(p+q)}{2 !}\left(\frac{\alpha}{q}\right)^2 \\
&+\frac{p(p+q)(p+2 q)}{3 !}\left(\frac{\alpha}{q}\right)^3+\ldots \ldots \text { (i) }
\end{aligned}
$$
On comparing Eqs. (i) and (ii), we get
$$
p=3, p+q=7, p+2 q=11
$$
and
$$
\frac{\alpha}{q}=\frac{1}{6} \Rightarrow q=4
$$
and
$$
\alpha=\frac{4}{6}=\frac{2}{3}
$$
So, let
$$
\begin{gathered}
x=(1-\alpha)^{-p / q}=\left(1-\frac{2}{3}\right)^{-3 / 4} \\
=\left(\frac{1}{3}\right)^{-3 / 4}=(3)^{3 / 4} \\
x^4=3^3=27
\end{gathered}
$$
$$
\therefore \quad x^4=3^3=27
$$
Hence, the correct answer is option: 'C'.